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A fair six-sided die is rolled five times. In the first three rolls, two of the outcomes were even. Similarly, in the last three rolls, two of the outcomes were even. What is the probability that the third roll resulted in an even number?
A. \(\frac{1}{2}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{4}{5}\)
E. \(\frac{5}{6}\)
A. \(\frac{1}{2}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{4}{5}\)
E. \(\frac{5}{6}\)
06 May 2024, 00:37
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Official Solution:
A fair six-sided die is rolled five times. In the first three rolls, two of the outcomes were even. Similarly, in the last three rolls, two of the outcomes were even. What is the probability that the third roll resulted in an even number?
A. \(\frac{1}{2}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{4}{5}\)
E. \(\frac{5}{6}\)
The sequence where the third roll is NOT even, while two of the outcomes in the first three rolls are even and two of the outcomes in the last three rolls are also even is only EEOEE.
The sequences in which the third roll IS even, while two of the outcomes in the first three rolls are even and two of the outcomes in the last three rolls are also even are:
EOEEO
EOEOE
OEEEO
OEEOE
Therefore, the condition given in the question is satisfied in 4 out of 5 possible sequences, yielding a probability of \(\frac{4}{5}\).
Answer: D
A fair six-sided die is rolled five times. In the first three rolls, two of the outcomes were even. Similarly, in the last three rolls, two of the outcomes were even. What is the probability that the third roll resulted in an even number?
A. \(\frac{1}{2}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{4}{5}\)
E. \(\frac{5}{6}\)
The sequence where the third roll is NOT even, while two of the outcomes in the first three rolls are even and two of the outcomes in the last three rolls are also even is only EEOEE.
The sequences in which the third roll IS even, while two of the outcomes in the first three rolls are even and two of the outcomes in the last three rolls are also even are:
EOEEO
EOEOE
OEEEO
OEEOE
Therefore, the condition given in the question is satisfied in 4 out of 5 possible sequences, yielding a probability of \(\frac{4}{5}\).
Answer: D
29 May 2024, 02:22
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I think this is a high-quality question and I agree with explanation. Is there another method to this rather than using E and O and write all the possible outcomes
